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BITSAT 2012 Paper

Section: Mathematics

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Question : 104 of 150

Marks: +1, -0

If f(x) =

x^\alpha

log x and f(0) = 0, then the value of α for which Rolle’s theorem can be applied in [0, 1] is

–2
–1
0
1/2

Solution:

For Rolle’s theorem in [a, b], f(a) = f(b),

In [0, 1] ⇒ f(0) = f(1) = 0

Since the function has to be continuous in [0, 1]

⇒ f (0) =

\lim\limits_{x\to 0^{+}}

f (x) = 0

⇒

\lim\limits_{x\to 0^{-}}

x^\alpha

log x = 0 ⇒

\lim\limits_{x\to 0}

\frac{\log x}{x^{-\alpha}}

= 0

Applying L.H. Rule

\lim\limits_{x\to 0}

\frac{\frac{1}{x}}{-\alpha x^{-\alpha-1}}

= 0

⇒

\lim\limits_{x\to 0}

\frac{-x^\alpha}{\alpha}

= 0 ⇒ α > 0

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